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Someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi. We think the actual amount is lower than that and want to run the test at an alpha level of 5%. What would our sample size need to be if we want to reject the null hypothesis if the sample mean is at or below 1,997.2956?

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7 votes

Answer:

The sample size must be greater than 37 if we want to reject the null hypothesis.

Explanation:

We are given that someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi.

Also, we are given a level of significance of 5%.

Let
\mu = mean breaking strength of their climbing rope

SO, Null Hypothesis,
H_0 :
\mu = 2,000 psi {means that the mean breaking strength of their climbing rope is 2,000 psi}

Alternate Hypothesis,
H_A :
\mu < 2,000 psi {means that the mean breaking strength of their climbing rope is lower than 2,000 psi}

Now, the test statistics that we will use here is One-sample z-test statistics as we know about population standard deviation;

T.S. =
(\bar X-\mu)/((\sigma)/(√(n) ) ) ~ N(0,1)

where,
\bar X = ample mean strength = 1,997.2956 psi


\sigma = population standard devaition = 10 psi

n = sample size

Now, at the 5% level of significance, the z table gives a critical value of -1.645 for the left-tailed test.

So, to reject our null hypothesis our test statistics must be less than -1.645 as only then we have sufficient evidence to reject our null hypothesis.

SO, T.S. < -1.645 {then reject null hypothesis}


(\bar X-\mu)/((\sigma)/(√(n) ) ) < -1.645


(1,997.2956-2,000)/((10)/(√(n) ) ) < -1.645


((1,997.2956-2,000)/(10)) * {√(n) } } < -1.645


-0.27044 * √(n)< -1.645


√(n)> (-1.645)/(-0.27044)


√(n)>6.083

n > 36.99 ≈ 37.

SO, the sample size must be greater than 37 if we want to reject the null hypothesis.

User Kevin Simper
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